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3 years ago

The law of reflection says that the angle of incidence is

1 answer:

4 0

The law of reflection states that the angle of incidence is equal to the angle of reflection. Furthermore, the law of reflection states that the incident ray, the reflected ray and the normal all lie in the same plane.

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Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

Which statement best describes the common features of sound waves and light waves

Viktor [21]

Sound waves or bounces off the wall and light waves are waves of light

7 0

3 years ago

What is the mechanical energy of a 500kg rollercoaster car moving with a speed of 3m/s at the top hill that is 30m high

koban [17]

K.E = 1/2*m*v^2 = 1/2(500)(3)^2 = 2250 J

m*g*h = 500(9.8)(30) = 147000 J

2250 + 147000 = 149250

4 0

3 years ago

A truck moves 30 km West, and then 40 km North, and then travels in a straight path back to its starting

SpyIntel [72]

Distance = (30+40+50) = 120 km

It's back where it started, so displacement = zero

7 0

3 years ago

A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is

Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2$

The acceleration of the craft should be 1.02234 m/s²

$F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N$

Weight of the craft

$W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N$

Thrust

$F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N$

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

3 0

3 years ago